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Svetlanka [38]
3 years ago
11

If the equation CO(g) + 2H2(g) →← → ← CH3OH(g) + energy is for a system at equilibrium, increasing the temperature will cause __

___. [
CH3OH] to increase and [CO] to decrease [

CH3OH] to decrease and [CO] to increase

both [CH3OH] and [CO] to increase

both [CH3OH] and [H2] to decrease
Chemistry
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

[CH₃OH] to decrease and [CO] to increase.

Explanation:

  • Since the energy appears as a product. So, the system is exothermic that releases heat.
  • Increasing the temperature of the system will cause the system to be shifted to the left side to attain the equilibrium again.
  • So, the right answer is:

<em>[CH₃OH] to decrease and [CO] to increase.</em>

<em></em>

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Which unstable element is used to determine the age of volcanic rock?
77julia77 [94]

<em><u>Answer:</u></em>

Potassium.

<u><em>Explanation:</em></u>

Therefore, the answer is Potassium. You might think, that because we were talking about Argon as well, the answer is both of them, but no. Everything starts with Potassium but it decays into Argon during the process.

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“If you start with a faster forward reaction, you will end up with more __________.”
Finger [1]

Answer:

time to do another things !

Explanation:

Here we can not fill in the blank the words like: products, because in many cases, a slower forward reaction can give the better quantity of products.

We can not say that a faster forward reaction gives the products with better quality, because it depend on what are the reactions we mention about.

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3 years ago
A 0.250 g sample of hydrocarbon (containing only carbon and hydrogen) undergoes complete combustion to produce 0.845 g of CO2 an
melomori [17]

Answer:

CH

Explanation:

We have to obtain the mass of carbon and hydrogen in CO2 and H2O respectively.

For carbon in CO2;

0.845 * 12/44 = 0.23 g

For hydrogen in H20;

0.173 * 2/18 = 0.019 g

We convert the masses to moles of carbon and hydrogen

For carbon - 0.23/ 12 = 0.019 moles

For hydrogen - 0.019/1 = 0.019 moles

Dividing by 0.019 moles for carbon and hydrogen we have the emperical formula of the compound as CH

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3 years ago
24 POINTS!!!!!!!!
Mama L [17]
B. A bond between two atoms: it doesn't matter if it's positive or negative or neutral, if there is a bond between two atoms, it is covalent.
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3 years ago
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A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f
jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is N_4O_3. Empirical formula is the smallest whole number ratios of elements in a compound.  

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89

The moles may be written as the temporary subscripts of the elements in the compound as follows:

N_{3.85}O_{2.89}

STEP 4: Divide the moles by the smallest mole value.

N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\

<u>STEP 5:</u> Multiply the subscripts by a number that would give the smallest whole number ratio.  

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}

mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%

Since the values are the same from the given, the answer is correct.

LEARN MORE

  • Writing Chemical Formula brainly.com/question/4697698
  • Naming Compounds brainly.com/question/8968140
  • Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

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3 years ago
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